# Prove that the sum of the squares of the first n integer is 1/24 (2n) (2n+1) (2n+2).?

### 6 Answers

- seahLv 71 decade agoBest answer
Let say

1^2 + 2^2 + 3^2 + ... + n^2 = 1/24 (2n) (2n+1) (2n+2)

is true for n

for n+1,

1^2 + 2^2 + 3^2 + ... + n^2 + (n+1)^2

= 1/24 (2n) (2n+1) (2n+2) + (n+1)^2

= 1/24 (2n) (2n+1) (2n+2) + 1/24 (24)(n+1)^2

= 1/24 [(2n) (2n+1) (2n+2) + 24(n+1)^2]

= 1/24 [(2n)(4n^2+6n+2) + 24(n^2+2n+1)]

= 1/24[8n^3 +12n^2 +4n + 24n^2 +48n + 24]

= 1/24[8n^3 + 36n^2 +52n + 24]

= 1/24 ((2n+2)(2n+3)(2n+4)

= 1/24 (2(n+1)) (2(n+1)+1)(2(n+1)+2)

Proved for n+1

for n= 1

RHS

1/24 (2(1)) (2(1)+1) (2(1)+2)

= 1/24 (2)(3)(4)

= 1 = 1^2

same as LHS

So, use Inductiom method

1^2 + 2^2 + 3^2 + ... + n^2 = 1/24 (2n) (2n+1) (2n+2)

is true for n >=1

Source(s): myseah - RackbraneLv 71 decade ago
Start from the identities:

(k+1)^2 - k^2 = 2k + 1 ........(1)

(k+1)^3 - k^3 = 3k^2 + 3k + 1 ..........(2)

Then:

(2) * 2 - (1) * 3 gives:

2[ (k+1)^3 - k^3 ] - 3[ (k+1)^2 - k^2 ]

= 6k^2 - 1 .........(3)

Note that when summing the left hand side from k = 1 to k = n, all terms in each of the square brackets disappear, except for the first term when k = n and the last term when n = 1.

If s is the sum you require, then summing (3)

from k = 1 to k = n gives:

6s - 1

= 2[ (n + 1)^3 - 1 ] - 3[ (n + 1)^2 - 1 ]

= 2(n + 1)^3 - 3(n +1)^2 + 1

6s = 2(n + 1)^3 - 3(n +1)^2 + 2

= 2n^3 + 6n^2 + 6n + 2 - 3n^2 - 6n + 1

= 2n^3 + 3n^2 + 1

= n(2n + 1)(n + 1)

Multiplying by 4 to obtain the rather unusual form you have:

24s = (2n)(2n + 1)(2n + 2)

s = (1/24)(2n)(2n + 1)(2n + 2).

- 3 years ago
start up from the identities: (ok+a million)^2 - ok^2 = 2k + a million ........(a million) (ok+a million)^3 - ok^3 = 3k^2 + 3k + a million ..........(2) Then: (2) * 2 - (a million) * 3 provides: 2[ (ok+a million)^3 - ok^3 ] - 3[ (ok+a million)^2 - ok^2 ] = 6k^2 - a million .........(3) be conscious that when summing the left hand facet from ok = a million to ok = n, all words in all the sq. brackets disappear, apart from the 1st time era whilst ok = n and the final time era whilst n = a million. If s is the sum you require, then summing (3) from ok = a million to ok = n provides: 6s - a million = 2[ (n + a million)^3 - a million ] - 3[ (n + a million)^2 - a million ] = 2(n + a million)^3 - 3(n +a million)^2 + a million 6s = 2(n + a million)^3 - 3(n +a million)^2 + 2 = 2n^3 + 6n^2 + 6n + 2 - 3n^2 - 6n + a million = 2n^3 + 3n^2 + a million = n(2n + a million)(n + a million) Multiplying via 4 to acquire the truly unusual style you have: 24s = (2n)(2n + a million)(2n + 2) s = (a million/24)(2n)(2n + a million)(2n + 2).

- 1 decade ago
First of all,

2n ( 2n + 1) ( 2n + 2) / 24 = n ( n + 1) ( 2n + 1) / 6 .

We want to prove :

Σ i² = 1² + 2² + 3² + . . . + n² = n ( n + 1) ( 2n + 1) / 6 .

If true for n = r , then adding ( r + 1)² to both sides

1² + 2² + 3² + . . . + r² + ( r + 1)² = ( r + 1) ( r + 2) ( r + 3) / 6 .

Substituting r + 1 for n shows that n + 1 is true,

and since 1² = 1 ( 1+ 1) ( 2 + 1) / 6 is true,

then every case for n is true.

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- smciLv 71 decade ago
To prove: Σi² = (n)(2n+1)(n+1)/6

Generally you either prove it by induction.

(I was thinking you could prove it directly by squaring the more basic result of the sum of the first n integers:

Σi = (n)(n+1)/2 = 1 +2 + ... + n

but apparently you can't)

Inductive proof:

Assume Σi²_i=1_(n-1) = (n-1)(2n-1)(n)/6

Then Σi²_i=1_n = Σi²_i=1_(n-1) + n²

= (n-1)(2n-1)(n)/6 + n²

= [ (n-1)(2n-1)(n) + 6n² ]/6

= [ (n-1)(2n-1) + 6n ] * n/6

= [ 2n²-3n+1 + 6n ] * n/6

= [ 2n²+3n+1 ] * n/6

= (n+1)(2n+1)n/6

QED.

- yasiru89Lv 61 decade ago
Let me present something very general, never let people get away with induction as if they derived it,

consider,

f(n) = n^r .(n+1)^r - n^r .(n-1)^r -----(1)

f(1) + f(2) +...+f(n) = n^r .(n+1)^r

and the binomial theorem gives,

f(n) /2 = rC1.n^(2r-1) + rC3.n^(2r-3)+....

putting (n-1),(n-2),...,2,1 in succession for n and adding we have,

n^r.(n+1)^r /2 = rC1 S(2r-1) + rC3 S(2r-3) +...

where S(2r -k) is a sum of odd powers of integers.

substitute a suitable r and get S1,S2,etc.

Now find a similar relation for sums of even powers of integers, all you have to do is change a negative sign in (1)